Problem: $\lim_{x\to 4}\dfrac{-5x^2+20x}{x^3-3x^2-4x}=$
Answer: Substituting $x=4$ into $\dfrac{-5x^2+20x}{x^3-3x^2-4x}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{-5x^2+20x}{x^3-3x^2-4x}$ can be simplified as $\dfrac{-5}{x+1}$, for $x\neq 0,4$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $0$ and $4$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{-5x^2+20x}{x^3-3x^2-4x}=\dfrac{-5}{x+1}$ for all $x$ -values in the interval $(3,5)$ except for $x=4$. Therefore, $\lim_{x\to 4}\dfrac{-5x^2+20x}{x^3-3x^2-4x}=\lim_{x\to 4}\dfrac{-5}{x+1}=-1$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 4}\dfrac{-5x^2+20x}{x^3-3x^2-4x}=-1$.